If x in(-pi, pi) then the number of solutions | vedclass

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(C) Given equation: $2 \sin x \sin 3 x \sin 5 x + \sin 5 x \cos 4 x = 0$Factor out $\sin 5 x$: $\sin 5 x (2 \sin x \sin 3 x + \cos 4 x) = 0$Using the

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$13$

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(C) Given equation: $2 \sin x \sin 3 x \sin 5 x + \sin 5 x \cos 4 x = 0$Factor out $\sin 5 x$: $\sin 5 x (2 \sin x \sin 3 x + \cos 4 x) = 0$Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$2 \sin x \sin 3 x = \cos(x-3x) - \cos(x+3x) = \cos(-2x) - \cos(4x) = \cos 2x - \cos 4x$Substitute back: $\sin 5 x (\cos 2x - \cos 4x + \cos 4x) = 0$$\sin 5 x \cos 2x = 0$This implies $\sin 5 x = 0$ or $\cos 2x = 0$.For $\sin 5 x = 0$,$5x = n\pi \implies x = \frac{n\pi}{5}$ for $n \in \mathbb{Z}$.In $(-\pi, \pi)$,$n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$,giving $9$ solutions.For $\cos 2x = 0$,$2x = (2k+1)\frac{\pi}{2} \implies x = (2k+1)\frac{\pi}{4}$ for $k \in \mathbb{Z}$.In $(-\pi, \pi)$,$k \in \{-2, -1, 0, 1\}$,giving $4$ solutions: $\pm \frac{\pi}{4}, \pm \frac{3\pi}{4}$.Total distinct solutions: $9 + 4 = 13$.

If $x \in(-\pi, \pi)$ then the number of solutions of the equation $2 \sin x \sin 3 x \sin 5 x+\sin 5 x \cos 4 x=0$ is

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